Lets consider polynomials of degree 4, with coefficients in $\mathbb{Q}$ and the ratios of their roots being roots of unity. Moreover, let those roots to be distinct. Now, consider the following given lists of roots: $$r^{1/4}(1,i,-1,-i), \quad r^{1/2}(\omega^2, -\omega^2,\omega,-\omega), \quad r(\epsilon, \epsilon^2, \epsilon^3, \epsilon^4), \quad r \sqrt2 (\xi \omega, \xi \omega^2, \xi^{-1}\omega, \xi^{-1}\omega^2)$$ where $\epsilon = exp(2 \pi i/5), \omega=exp(2 \pi i/3), \xi=exp(\pi i /4)$ and $r \in \mathbb{Q}$.
My question now is, how i can find all those polynomials, whose roots are listed above?
My idea:
Lets take the second one for example. So let $p(x) \in \mathbb{Q}[x]$ be a polynomial of degree 4 and its roots are $r^{1/2}(\omega^2, -\omega^2,\omega,-\omega)$. If we know the roots of $p$, we also know its factors, i.e. $$p(x) = (x - r^{1/2}\omega^2)(x + r^{1/2}\omega^2)(x - r^{1/2}\omega)(x + r^{1/2}\omega).$$ Multiplying this out, we get $$p(x) = (r \omega^2 - x^2)(r \omega^4 - x^2) = x^4 - r \omega^2 (1 + w^2)x^2 + r^2 \omega^6.$$ As in the comment below noted, applying $\omega^3=1, 1+\omega+\omega^2=0$ leaves us with $- r \omega^2 (1 + (-1-\omega)) = r \omega^3 = r$ for thequadratic coefficient. So finally we get $$p(x) = x^4 + r x^2 + r^2.$$